AB is tangent to circle O at B. Find the length of the radius r for AB = 5 and AO = 8.6. Round to the nearest tenth if necessary
2 answers:
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Answer: The answers are (a) 40 cm and (b)
Step-by-step explanation: The calculations are as follows:
(a) See the figure (a). As given in the question, A circle with centre 'O' circumscribes a triangle ABC with BC = 20 cm and ∠BAC = 30°. We need to find the diameter DC of the circle.
Let us draw BD. Now, ∠BAC and ∠BDC are angles on the same arc BC, so we have
∠BAC = ∠BDC = 30°.
Also, ∠CBD = 90°, since it stands on the diameter DC. So, ΔBCD will be a right angled triangle.
We can write
Thus, the diameter of the circle = 40 cm.
(b) See the figure (b).
As given in the question, A circle with centre 'O'' circumscribes a triangle DEF with EF = 4.6 inches and diameter GF = 12.4 in.. We need to find the angle EDF.
Let us draw GE. Now, ∠EGF and ∠EDF are angles on the same arc EF, so we have
∠EGF = ∠EDF = ?
Also, ∠GEF = 90°, since it stands on the diameter GF. So, ΔGEF will be a right angled triangle.
We can write
Thus,
<u><em>Answer:</em></u>
The length of the diagonal is 284 km
<u><em>Explanation:</em></u>
<u>1- we get the side length of the side:</u>
We are given that the land has a square shape and an area of 40,404 km²
Area of square = (side)²
40,404 = (side)²
side = +√40,404
side = 201.007 km ≈ 201 km
<u>2- we get the length of the diagonal:</u>
Refer to the attach picture representing the area of land.
We can infer that the two side lengths along with the diagonal of the land form a right-angled triangle.
Therefore, to get the diagonal, we can apply the Pythagorean theorem as follows:
Diagonal = 284.2 ≈ 284 km
Hope this helps :)
